For every real number x there exists an integer $n$ such that $n>x$.
The book is using contradiction,
Suppose $x$ is a real number such that $n≤x$ for every $n$,that mean $x$ is the upper bound of natural number set.
then $M$ is an lowest upper bound of the set of integers and i guess that M is the integer.
$M-1<M$,meaning that $M-1$ must not the upper bound.
there will have $N>M-1$ ,because $M-1$ not the upper bound so there is a natural number N greater than it
$N+1>M$,that mean always have the integer greater than the lowest upper bound of the set of integers,contridcate with natural number set having lowest upper bound.
the above only show that the natural number is unbounded by natural number,how is it related to natural number is also unbounded by the real number?
Also,i watched other proof on the website,it is curious that it derived $m+1>x$ in the conclusion. I mean if i take $x=8.5$,natural number set$=\{0,1,2,3,4,5,6\}$,then the lub of this set will be 6,and 5 is not upper bound of the this set,then m+1 will be 7,and 7 is not belong to set S but $(m+1>x)$ 7 is not greater than 8.5.
